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100 mg mass of nichrome metal is drawn into a wire of area of cross-section 0.05 mm. Calculate the resistance of this wire. Given density of nichrome 8.4 xx 10^3 "kgm"^(-3)and resistivity of the material as 1.2 xx 10^(-6) Omegam. |
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Answer» Solution :Given mass = 100 mg =`10^2xx10^(-6)=10^(-4)` kg area of CROSS section A = `5xx10^(-2) xx (10^(-3))^2m^2` i.e.,`A=5xx10^(-8)m^2`. DENSITY`D=8.4xx10^3 kgm^(-3), rho=1.2xx10^(-6) Omegam` but density = `"mass"/"volume"` i.e. volume = `"mass"/"density"=10^(-4)/(8.4xx10^3)` `=(0.119xx10^(-7))/(5xx10^*(-8))=0.0238xx10` i.e. length= 0.238 m We know that , `R=(rhol)/A =(1.2xx10^(-6)xx0.238)/(5xx10^(-8))` =0.05712 x 100 or R=5.712 `Omega` Resistanceof the wire = `5.712Omega`. |
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